Precalculus: Mathematics for Calculus, 7th Edition

by
Stewart, James; Redlin, Lothar; Watson, Saleem

Published by
Brooks Cole

ISBN 10:
1305071751

ISBN 13:
978-1-30507-175-9

Chapter 1 - Review - Exercises: 68

Answer

$x=\pm3$

Work Step by Step

$x^{4}-8x^{2}-9=0$
Let $x^{2}$ be equal to $u$
If $u=x^{2}$, then $u^{2}=x^{4}$.
Rewrite the original equation using the new variable $u$:
$u^{2}-8u-9=0$
Solve by factoring:
$(u-9)(u+1)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-9=0$
$u=9$
$u+1=0$
$u=-1$
Substitute $u$ back to $x^{2}$ and solve for $x$:
$u=9$
$x^{2}=9$
$x=\sqrt{9}$
$x=\pm3$
$u=-1$
$x^{2}=-1$
$x=\sqrt{-1}$
Since $x=\sqrt{-1}$ won't yield real solutions, the final answer is $x=\pm3$